HDU 1061 Rightmost Digit

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1061

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 39835 Accepted Submission(s): 15036

Problem Description

Given a positive integer N, you should output the most right digit of N^N.

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.

Each test case contains a single positive integer N(1<=N<=1,000,000,000).

Output

For each test case, you should output the rightmost digit of N^N.

Sample Input

2
3
4

Sample Output

7
6

Hint
In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7.
In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6.

Author
Ignatius.L

Recommend
大意——给你一个正整数n，要你算出n^n的最后一位数码。其中1<=n<=1,000,000,000。

思路——与HDU 1097差不多，我们可以用二分求快速幂，并且边乘边模，这样的话就不会出问题了。

复杂度分析——时间复杂度:O(log(n)),空间复杂度:O(1)

附上AC代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <cmath>
#include <iomanip>
#include <ctime>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>
using namespace std;
typedef unsigned int UI;
typedef long long LL;
typedef unsigned long long ULL;
typedef long double LD;
const double pi = acos(-1.0);
const double e = exp(1.0);
const double eps = 1e-8;
const int mod = 10;

int quick_pow(int a);

int main()
{
	ios::sync_with_stdio(false);
	int T, a;
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d", &a);
		printf("%d\n", quick_pow(a));
	}
	return 0;
}

int quick_pow(int a)
{
	int res = 1;
	int b = a;
	while (b > 0)
	{
		if (b & 1)
			res = ((res%mod)*(a%mod))%mod;
		a = ((a%mod)*(a%mod))%mod;
		b >>= 1;
	}
	return res;
}